Php mysql登录用户名密码

这是我用来检查用户是否与表中的密码（同一行）匹配的代码。 我得到一个错误：

警告：mysql_query（）：用户“apache”@“localhost”的访问被拒绝 （使用密码：否）在中 /主页/csc4370FA14_18/public_html/program/assignments/group 第14行的project3/login.php警告：mysql\u query（）：指向 无法在中建立服务器 /主页/csc4370FA14_18/public_html/program/assignments/group 第14行的project3/login.php警告：mysql\u num\u rows（）需要 参数1是资源，布尔值在 /主页/csc4370FA14_18/public_html/program/assignments/group 第15行的project3/login.php

我不知道为什么这可能是不正确的，因为登录凭据等工作正常。我认为这与mysqli有关，但与mysql_*函数相比，我不太了解这一点。事实上，我知道这是正确的连接信息。

试试这段代码

$servername = "localhost";
$username = "csc4370FA14_18";
$password = "1db23";
$dbname = "csc4370FA14_18";


$conn = mysqli_connect($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$username_login = $_POST["username"];
$password_login = $_POST["pw"];

$query2 = mysql_query("SELECT * FROM users WHERE name='$username_login'");
$numrow = mysql_num_rows($query2);

if ($numrow != 0) {
    while ($row = mysql_fetch_assoc($query2)) {
        $dbusername = $row['name'];
        $dbpassword = $row['password'];
    }

    // Check to see if username and password match
    if ($username_login==$dbusername && $password_login==$dbpassword) {
        echo "You are in";
    }
    else {
        echo "Sorry $username_login. Incorrect password!";
    }
}

---

这回答了你的问题吗？

$servername = "localhost";
$username = "csc4370FA14_18";
$password = "1db23";
$dbname = "csc4370FA14_18";


$conn = mysqli_connect($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$username_login = $_POST["username"];
$password_login = $_POST["pw"];

$query2 = mysqli_query($conn,"SELECT * FROM users WHERE name='$username_login'");
$numrow = mysqli_num_rows($query2);

if ($numrow != 0) {
    while ($row = mysqli_fetch_assoc($query2)) {
        $dbusername = $row['name'];
        $dbpassword = $row['password'];
    }

    // Check to see if username and password match
    if ($username_login==$dbusername && $password_login==$dbpassword) {
        echo "You are in";
    }
    else {
        echo "Sorry $username_login. Incorrect password!";
    }
}